Question

Let $f^{{}^{\prime }}\left(1\right)=2$ and $f\left(1\right)=4$, then the derivative of $\log \left[f\right(e^x\left)\right]$ at the point $x=0$ is equal to

• A. $2$
• B. $1$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

$h\left(x\right)=\log \left(f\right(e^x\left)\right)$
Then $h^{\prime }\left(x\right)=\frac{1}{f\left(e^x\right)}.f^{\prime }\left(e^x\right).e^x$
At $x=0,h^{\prime }\left(0\right)=\frac{1}{f\left(e^0\right)}.f^{\prime }\left(e^0\right).e^0$

$=\frac{1}{f\left(1\right)}.f^{\prime }\left(1\right).1$

$=\frac{f^{\prime }\left(1\right)}{f\left(1\right)}=\frac{2}{4}=\frac{1}{2}$